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Euclidian topology ang cofinite topology

  1. Oct 23, 2009 #1
    please can you help me to prove this exercise;
    Prove that:
    the Euclidean topology R is finer than the cofinite topology on R



    please answer me as faster as u can I have an exame on monday and I don't know to provethis exercise!
     
  2. jcsd
  3. Oct 23, 2009 #2
    Consider an arbitrary open set U in the cofinite topology. U is all of R except [itex]x_1,x_2,\ldots,x_n[/itex]. Given some point [itex]x \in U[/itex] can you find an open interval I containing x, but included in U. That is [itex]x \in I \subseteq U[/itex] where I is open in the standard Euclidean topology and U is open in the cofinite topology. (HINT: can you show that the open interval centered at x with radius [itex]\min(|x-x_1|,|x-x_2|,\ldots,|x-x_n|)[/itex] works?)

    It's a well-known theorem that if for every open set U in a topology T, and every point [itex]x\in U[/itex] we can find some set U' such that [itex]x \in U' \subseteq U[/itex] and U' is open in the topology T', then T' is finer than T.
     
  4. Oct 24, 2009 #3
    thank you! U had helped me a lot thanks thanks
     
  5. Oct 24, 2009 #4
    Can you help me and for this exercise please!

    the function f:(X,T)->(X,T') is continuous
    the function f:(X,T')->(X,T) is open



    {T is the Euclidean topology,T' is cofinite topology}
    [the function f maybe is the identity function id but I am not sure]
     
  6. Oct 24, 2009 #5
    I'm assuming that you're asked to prove the two statements.

    If f is the identity function then you're just restating the result in the OP. That f:(X,T)->(X,T') is continuous means that for every set U open in T', the set [itex]f^{-1}(U) = U[/itex] is open in T. Hence this is equivalent to "if U is open in T', then U is open in T" which is just stating that T is finer than T'. The other statement is shown in exactly the same way.

    If f is an arbitrary function then the result is not true because for every [itex]y \in X[/itex] we can let [itex]f_y : (X,T') \to (X,T)[/itex] be the constant function f(x) = y so [itex]f(X) = \{y\}[/itex] is open according to your second statement. Since every one-element set is open T is the discrete topology, but the Euclidean topology is different from the discrete topology (unless of course X consists solely of isolated points).
     
  7. Oct 24, 2009 #6
    Thank you!
     
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